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It depends on the concentration of HCl in the solution.
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How many moles of solute are in 50 mL of a 12 M HCl solution?
Molarity = moles of solute/Liters of solution ( 50 ml = 0.05 Liters ) 12 M HCl = moles HCl/0.05 Liters = 0.60 moles HCl
How many moles of CO2 form when 15.5 mL of 3.00M HCl solution react?
How many moles of CO2 form when 15.5 ml of 3.00m HCl solution react?
How many moles are dissolved in 200 ml of 0.100 M HCl?
20
How many moles of HCl are found in 50 mL of a 6.0 M solution?
50 mL = 0.05 L (6.0 moles/L) × (0.05 L) = 0.3 moles
How many moles of HCl are present in 40ml of a 0.035M solution?
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
How many moles of solute are in 50 mL of a 12 M HCl solution?
Molarity = moles of solute/Liters of solution ( 50 ml = 0.05 Liters ) 12 M HCl = moles HCl/0.05 Liters = 0.60 moles HCl
How many moles of CO2 form when 15.5 mL of 3.00M HCl solution react?
How many moles of CO2 form when 15.5 ml of 3.00m HCl solution react?
How many moles of HCl are in 3.36 mL of 22.22 M HCl and 81.5 mL of water?
0.0747mol/L of HCL 1.81mol/L of water
How many moles are dissolved in 200 ml of 0.100 M HCl?
20
How many moles of HCl are found in 50 mL of a 6.0 M solution?
50 mL = 0.05 L (6.0 moles/L) × (0.05 L) = 0.3 moles
How many moles of HCl are present in 40ml of a 0.035M solution?
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
How many moles of HCl are present in 75 mL of a 200 M solution?
.0150 mol
How many moles of sodium hydroxide are used for the tiltration if 17.65 mL of a 0.110 M sodium is needed to titrated 25.00 mL of a hydrochloride acid solution?
I need to see the balanced equation to work!NaOH + HCl --> NaCl + H2O ( good, all one to one )Now, find molarity HCl ( sodium, or sodium hydroxide; no matter )(17.65 mL)(0.110 M NaOH) = (25.00 mL)(X M HCl)= 0.07766 M HCl-------------------------now,Molarity = moles of solute/Liters of solution ( 25.00 mL = 0.025 L)0.07766 M HCl = X moles/0.025 Liters= 0.001942 moles HCl---------------------------------------formal set up, though not needed0.001942 moles HCl (1 mole NaOH/1 mole HCl)= 0.00194 moles sodium hydroxide used=============================
750 ml of 20M HCl is mixed with 250 ml of 60M HCl What is the new molarity?
Molarity of a solution is the number of moles of the solute divided by the volume of the solution (in liters). If 750 ml of 20M HCl is mixed with 250 ml of 60M HCl, we first find the total number of moles of HCl in our new solution. Using that same formula, M=moles/V, we cansee that moles=MV. In the first solution we have (20M)(0.750L) = 15 moles. In the second, (60M)(0.250L) = 15 moles, so we have a total of 30 moles in our new solution, which also has a volume of 750mL + 250mL = 1L. The molarity of the new solution is 30 moles/1L = 30M
How many moles of HCL gas are contained in 25 ml of a 0.10 M solution of the acid?
you cant solve that without the concentration of the HCl or the mass if you have either of these use: To find the moles with mass of HCl you have use n= m/mm , m=mass of HCl and mm = the molar mass so the elements mass combined from periodic table to find the moles with the volume = 0.06 (in litres must be!) and concentration use n = cv
What is the molarity of a solution prepared by dissolving 1.56 grams of gaseous HCl in enough water to make 26.8 mL of solution?
Find moles of HCl first. 1.56 grams HCl (1mole HCl/36.458 grams) = 0.0428 moles HCl Molarity = moles of solute/volume of solution Molarity = 0.0428 moles/26.8 ml = 0.00160 milli-Molarity, or more to the point; = 1.60 X 10^-6 Molarity of HCl
How many grams of sodium carbonate is needed to react with 18.1 ml of 0.23M HCL?
Balanced equation always first!! Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O Find moles HCl by---Molarity = moles of solute/Liters of solution (18.1 ml = 0.0181 L) 0.23 M HCl = moles HCl/0.0181 liters = 0.004163 moles HCl Drive back against sodium carbonate stoichometrically 0.004163 moles HCl (1 mole Na2CO3/2 mole HCl)(105.99 grams/1 mole Na2CO3) = 0.22 grams Na2CO3 needed --------------------------------------------
